If the roots of the equation $a\left(b - c\right)x^{2} + b\left(c - a\right)x + c\left(a - b\right) = 0$ are equal then prove that $2/b = 1/a + 1/c$, i.e. $a,b,c$ are on H.P. I have tried to prove this using the quadratic formula and also with the zero-coefficient relationship but at the end the expression becomes too cumbersome to solve.
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$x=1$ is clearly a root, and hence a double root. Product of roots then yields $c(a-b) = a(b-c)$, from which you can prove the desired result. – Hari Shankar Dec 19 '17 at 03:51
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Set $1/a=A,1/b=B,1/c=C$ to find
$$(C-B)x^2+(A-C)x+(B-A)=0$$
So, we need to establish $A,B,C$ are in A.P.
As the discriminant has to be $=0,$
$$0=(A-C)^2-4(C-B)(B-A)=(C-B+B-A)^2-4(C-B)(B-A)=?$$
lab bhattacharjee
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So it discriminat must be $0$:
$$ b^2(c-a)^2 = 4ac(a-b)(b-c)$$
Let $x=b/a$ and $z=b/c$. Then $$ (z-x)^2 = 4(x-1)(1-z)$$ so $$ x^2+z^2-2zx = 4z+4x-4-4zx$$ so $$ (x+z)^2 = 4(x+z)-4$$ thus $$(x+z-2)^2 =0$$ and we are done.
nonuser
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Painstakingly reached $$\left(\dfrac{b-c}{bc}-\dfrac{a-b}{ab}\right)^2=0$$ – lab bhattacharjee Dec 18 '17 at 15:41